Class 11 Maths NCERT Solutions Chapter 9 Ex 9.2

Every point on the x-axis has y-coordinate zero, and every point on the y-axis has x-coordinate zero.

Using the point-slope form with (x_0,y_0)=(-4,3) and m=\dfrac{1}{2}:

y-3=\dfrac{1}{2}(x-(-4))

2(y-3)=x+4\;\Rightarrow\;2y-6=x+4

x-2y+10=0

Using the point-slope form with (x_0,y_0)=(0,0):

y-0=m(x-0)

Slope of the line, using 75^\circ=45^\circ+30^\circ and the tangent addition formula:

m=\tan75^\circ=\tan(45^\circ+30^\circ)=\dfrac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ}

=\dfrac{1+\dfrac{1}{\sqrt{3}}}{1-\dfrac{1}{\sqrt{3}}}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}

Rationalising the denominator:

m=\dfrac{(\sqrt{3}+1)^2}{(\sqrt{3}-1)(\sqrt{3}+1)}=\dfrac{3+2\sqrt{3}+1}{3-1}=\dfrac{4+2\sqrt{3}}{2}=2+\sqrt{3}

Using the point-slope form with (x_0,y_0)=(2,2\sqrt{3}) and m=2+\sqrt{3}:

y-2\sqrt{3}=(2+\sqrt{3})(x-2)

y-2\sqrt{3}=(2+\sqrt{3})x-2(2+\sqrt{3})

y-2\sqrt{3}=(2+\sqrt{3})x-4-2\sqrt{3}

(2+\sqrt{3})x-y-4=0

Since the line meets the x-axis at a distance of 3 units to the left of the origin, its x-intercept is d=-3. Using the slope-intercept form for x-intercept, y=m(x-d), with m=-2:

y=-2(x-(-3))=-2(x+3)

y=-2x-6

2x+y+6=0

The y-intercept is c=2, and the slope is m=\tan30^\circ=\dfrac{1}{\sqrt{3}}. Using the slope-intercept form y=mx+c:

y=\dfrac{1}{\sqrt{3}}x+2

Multiplying throughout by \sqrt{3}:

\sqrt{3}\,y=x+2\sqrt{3}

x-\sqrt{3}\,y+2\sqrt{3}=0

Slope of the line through (-1,1) and (2,-4):

m=\dfrac{-4-1}{2-(-1)}=\dfrac{-5}{3}

Using the two-point form with (x_1,y_1)=(-1,1):

y-1=-\dfrac{5}{3}(x-(-1))

3(y-1)=-5(x+1)

3y-3=-5x-5

5x+3y+2=0

The median through R passes through R and the mid-point of the opposite side PQ. Mid-point of PQ:

M=\left(\dfrac{2+(-2)}{2},\dfrac{1+3}{2}\right)=(0,2)

Slope of the median through R(4,5) and M(0,2):

m=\dfrac{5-2}{4-0}=\dfrac{3}{4}

Using the point-slope form through R(4,5):

y-5=\dfrac{3}{4}(x-4)

4(y-5)=3(x-4)

4y-20=3x-12

3x-4y+8=0

Slope of the line through (2,5) and (-3,6):

m_1=\dfrac{6-5}{-3-2}=\dfrac{1}{-5}=-\dfrac{1}{5}

Since the required line is perpendicular to this line, its slope is the negative reciprocal:

m=-\dfrac{1}{m_1}=5

Using the point-slope form through (-3,5) with m=5:

y-5=5(x-(-3))

y-5=5x+15

5x-y+20=0

Let the point of division be D, dividing the segment joining A(1,0) and B(2,3) internally in the ratio 1:n. By the section formula:

D=\left(\dfrac{1(2)+n(1)}{1+n},\dfrac{1(3)+n(0)}{1+n}\right)=\left(\dfrac{n+2}{n+1},\dfrac{3}{n+1}\right)

Slope of AB:

m_{AB}=\dfrac{3-0}{2-1}=3

Since the required line is perpendicular to AB, its slope is:

m=-\dfrac{1}{3}

Using the point-slope form through D with m=-\dfrac{1}{3}:

y-\dfrac{3}{n+1}=-\dfrac{1}{3}\left(x-\dfrac{n+2}{n+1}\right)

Multiplying throughout by 3(n+1) and simplifying:

3(n+1)y-9=-(n+1)x+(n+2)

(n+1)x+3(n+1)y=n+11

Let the equal intercepts on both axes be a. Using the intercept form \dfrac{x}{a}+\dfrac{y}{a}=1:

x+y=a

Since the line passes through (2,3):

2+3=a\;\Rightarrow\;a=5

Let the x-intercept be a and the y-intercept be b, with a+b=9, i.e., b=9-a. Using the intercept form \dfrac{x}{a}+\dfrac{y}{b}=1 and substituting the point (2,2):

\dfrac{2}{a}+\dfrac{2}{9-a}=1

Multiplying throughout by a(9-a):

2(9-a)+2a=a(9-a)

18-2a+2a=9a-a^2

18=9a-a^2\;\Rightarrow\;a^2-9a+18=0

(a-3)(a-6)=0\;\Rightarrow\;a=3\text{ or }a=6

Case 1: a = 3, b = 6

\dfrac{x}{3}+\dfrac{y}{6}=1\;\Rightarrow\;2x+y=6

Case 2: a = 6, b = 3

\dfrac{x}{6}+\dfrac{y}{3}=1\;\Rightarrow\;x+2y=6

Slope of the line, with inclination \theta=\dfrac{2\pi}{3}=120^\circ:

m=\tan120^\circ=\tan(180^\circ-60^\circ)=-\tan60^\circ=-\sqrt{3}

Line through (0, 2)

Here the y-intercept is c=2. Using the slope-intercept form y=mx+c:

y=-\sqrt{3}\,x+2

\sqrt{3}\,x+y-2=0

Parallel line, 2 units below the origin

A parallel line has the same slope m=-\sqrt{3}. Since it crosses the y-axis 2 units below the origin, its y-intercept is c=-2:

y=-\sqrt{3}\,x-2

\sqrt{3}\,x+y+2=0

Slope of the segment joining the origin O(0,0) and the foot of the perpendicular P(-2,9):

m_{OP}=\dfrac{9-0}{-2-0}=-\dfrac{9}{2}

Since the required line is perpendicular to OP, its slope is the negative reciprocal:

m=-\dfrac{1}{m_{OP}}=\dfrac{2}{9}

Using the point-slope form through (-2,9):

y-9=\dfrac{2}{9}(x-(-2))

9(y-9)=2(x+2)

9y-81=2x+4

2x-9y+85=0

Since L is a linear function of C, treat (C,L) as coordinates of points on a line: (20,124.942) and (110,125.134). Slope of this line:

m=\dfrac{125.134-124.942}{110-20}=\dfrac{0.192}{90}=\dfrac{4}{1875}

Using the point-slope form through (20,124.942):

L-124.942=\dfrac{4}{1875}(C-20)

L=124.942+\dfrac{4}{1875}C-\dfrac{4}{1875}(20)

L=\dfrac{4}{1875}C+124.8993

Treating price P and demand D as coordinates of points on a line: (14,980) and (16,1220). Slope of this line:

m=\dfrac{1220-980}{16-14}=\dfrac{240}{2}=120

Using the point-slope form through (14,980):

D-980=120(P-14)

D=120P-1680+980=120P-700

At P=17:

D=120(17)-700=2040-700=1340

Let the line meet the x-axis at A(p,0) and the y-axis at B(0,q). Since P(a,b) is the mid-point of AB:

a=\dfrac{p+0}{2}\;\Rightarrow\;p=2a,\qquad b=\dfrac{0+q}{2}\;\Rightarrow\;q=2b

Using the intercept form of the equation of a line with x-intercept 2a and y-intercept 2b:

\dfrac{x}{2a}+\dfrac{y}{2b}=1

Multiplying both sides by 2:

\dfrac{x}{a}+\dfrac{y}{b}=2

Let the line meet the x-axis at A(p,0) and the y-axis at B(0,q). Since R(h,k) divides AB internally in the ratio 1:2 (from A to B), by the section formula:

h=\dfrac{1(0)+2(p)}{1+2}=\dfrac{2p}{3},\qquad k=\dfrac{1(q)+2(0)}{1+2}=\dfrac{q}{3}

Solving for p and q:

p=\dfrac{3h}{2},\qquad q=3k

Using the intercept form of the equation of a line:

\dfrac{x}{p}+\dfrac{y}{q}=1\;\Rightarrow\;\dfrac{x}{\dfrac{3h}{2}}+\dfrac{y}{3k}=1

\dfrac{2x}{3h}+\dfrac{y}{3k}=1

Multiplying both sides by 3:

\dfrac{2x}{h}+\dfrac{y}{k}=3

Let A(3,0), B(-2,-2) and C(8,2). Find the equation of the line through A and B, and check whether C satisfies it.

Slope of AB:

m=\dfrac{-2-0}{-2-3}=\dfrac{-2}{-5}=\dfrac{2}{5}

Using the point-slope form through A(3,0):

y-0=\dfrac{2}{5}(x-3)

5y=2x-6\;\Rightarrow\;2x-5y-6=0

Substituting C(8,2) into this equation:

2(8)-5(2)-6=16-10-6=0

Since C(8,2) satisfies the equation of the line through A and B, all three points lie on the same line.